Speed of the light and dilation of time

In summary: AND would the time on my watch show that 30 years have passed or something much less?In summary, if you were 35 years old when you got on the train and your son was 5 years old, and you traveled at near light speed for 30 Earth years, when you reunited with your son, you would be older than him. The total number of revolutions around Earth would be the same for both of you, but you would experience less time during each revolution, so you would age less. Your watch would also be behind your son's watch.
  • #176
GAsahi said:
Of course that the observer in circular motion is not seeing the same amount of redshift, since he isn't perfectly equivalent to the inertial observer. The point is that BOTH observers measure redshift.
OK, so now you have agreed that after A makes one complete loop around and has rejoined B, they both have observed the same redshift but A stops seeing it as soon as he stops but B continues to see the redshift for a longer period of time, in fact, the period of time that it takes for light to travel around the fiberoptic loop.

Now I don't want to cheat or be accused of moving any goalposts or even wanting to move any goalposts but I just don't have any idea how you make the transition to the actual scenario where A does not stop but rather passes by B and continues around the loop a second time. Could you please explain how the redshift works for both A and B and how it correlates to the actual time the A and B can see on each others clocks as they A passes by B?
 
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  • #177
ghwellsjr said:
OK, so now you have agreed that after A makes one complete loop around and has rejoined B, they both have observed the same redshift but A stops seeing it as soon as he stops but B continues to see the redshift for a longer period of time, in fact, the period of time that it takes for light to travel around the fiberoptic loop.

The above means that you realized your mistake in claiming that one observer measures redshift while the other measures blueshift. This is progress.

Now I don't want to cheat or be accused of moving any goalposts or even wanting to move any goalposts but I just don't have any idea how you make the transition to the actual scenario where A does not stop but rather passes by B and continues around the loop a second time. Could you please explain how the redshift works for both A and B and how it correlates to the actual time the A and B can see on each others clocks as they A passes by B?

Huh? The situation repeats identically. The period of the phaenomenon is [itex]2 \pi[/itex]. Both A and B measure redshift as soon as A starts accelerating away from B.
 
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  • #178
GAsahi said:
The above means that you realized your mistake in claiming that one observer measures redshift while the other measures blueshift. This is progress.
But it's only a tiny bit of progress. I still have a long way to go to learning all the goalpost rules. And thank you for not accusing me of trying to cheat this time. I really don't want any cheating going on.
GAsahi said:
Huh? The situation repeats identically. The period of the phaenomenon is [itex]2 \pi[/itex]. Both A and B measure redshift as soon as A starts accelerating away from B.
I see, so you're say that A stops when he reaches B so that B can finish seeing the redshift coming from A traveling all the way around the fiberoptic loop and then A accelerates away from B a second time. Now I realize that as soon as A stops, the redshift that he sees coming from B also immediately stops (no more Doppler) but how does A know when B stops seeing the redshift so that he can start up again? Is he sneaking a peek at B's webcam? Wouldn't that be cheating?
 
  • #179
ghwellsjr said:
But it's only a tiny bit of progress.

Yes, you are a slow learner.

I still have a long way to go to learning all the goalpost rules. And thank you for not accusing me of trying to cheat this time. I really don't want any cheating going on.

I see, so you're say that A stops when he reaches B so that B can finish seeing the redshift coming from A traveling all the way around the fiberoptic loop and then A accelerates away from B a second time. Now I realize that as soon as A stops, the redshift that he sees coming from B also immediately stops (no more Doppler) but how does A know when B stops seeing the redshift so that he can start up again?

A goes around in circles, nothing to do with B.
Is he sneaking a peek at B's webcam? Wouldn't that be cheating?

Where is this going? Are you just trolling now?
 
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  • #180
GAsahi said:
Yes, you are a slow learner.
If you quit teaching, then I'll never learn.
GAsahi said:
Where is this going?
I have no idea where this is going. You have all the rules and until you disclose them, I can't know them.
GAsahi said:
Are you just trolling now?
Not me, must be someone else.
 
  • #181
GAsahi said:
A goes around in circles, nothing to do with B.
Oops, you edited this in while I was composing my previous response.

So have you changed your mind, does A continue past B without stopping? I'm confused because before you said:
GAsahi said:
Huh? The situation repeats identically. The period of the phaenomenon is [itex]2 \pi[/itex]. Both A and B measure redshift as soon as A starts accelerating away from B.
That "Huh?" in there implies that something is obvious to you that you think should be obvious to me but it isn't, I just don't understand what you have in your mind, you need to spell it out, don't take anything for granted.
 
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  • #182
GAsahi said:
Huh? The situation repeats identically. The period of the phaenomenon is 2∏. Both A and B measure redshift as soon as A starts accelerating away from B.
ghwellsjr said:
That "Huh?" in there implies that something is obvious to you that you think should be obvious to me but it isn't, I just don't understand what you have in your mind, you need to spell it out, don't take anything for granted.
I have been going back over all of your posts and I think maybe I have figured out what you have been saying all along and I have been missing. Tell me if this is right:

The OP in this thread asked about a father-son variant of the twins paradox in which the father took a 30-year long trip at a very high speed and after he returned, his son had aged 30 years while he aged only a few days. In the typical twins paradox scenario, each observer always determines the other observer's clock is time dilated, both going and coming and yet when they reunite, the traveling observer has aged less. So whether the observed Doppler is red shifted (as at the beginning of the trip) or blue shifted (as at the end of the trip), they both lead to the same time dilation (clocks running slow), never time contraction (clocks running fast). The difference between the typical twins paradox and this one is that instead of the path being straight out and back, the path follows a circle.
 
  • #183
ghwellsjr said:
I have been going back over all of your posts and I think maybe I have figured out what you have been saying all along and I have been missing. Tell me if this is right:

The OP in this thread asked about a father-son variant of the twins paradox in which the father took a 30-year long trip at a very high speed and after he returned, his son had aged 30 years while he aged only a few days. In the typical twins paradox scenario, each observer always determines the other observer's clock is time dilated, both going and coming and yet when they reunite, the traveling observer has aged less. So whether the observed Doppler is red shifted (as at the beginning of the trip) or blue shifted (as at the end of the trip), they both lead to the same time dilation (clocks running slow), never time contraction (clocks running fast). The difference between the typical twins paradox and this one is that instead of the path being straight out and back, the path follows a circle.

Does this mean you have changed your mind about your conclusions regarding the direct communication scenario. either a transparent Earth or no earth?
 
  • #184
ghwellsjr said:
I have been going back over all of your posts and I think maybe I have figured out what you have been saying all along and I have been missing. Tell me if this is right:

The OP in this thread asked about a father-son variant of the twins paradox in which the father took a 30-year long trip at a very high speed and after he returned, his son had aged 30 years while he aged only a few days. In the typical twins paradox scenario, each observer always determines the other observer's clock is time dilated, both going and coming and yet when they reunite, the traveling observer has aged less. So whether the observed Doppler is red shifted (as at the beginning of the trip) or blue shifted (as at the end of the trip), they both lead to the same time dilation (clocks running slow), never time contraction (clocks running fast). The difference between the typical twins paradox and this one is that instead of the path being straight out and back, the path follows a circle.

Actually, I disagree with this analysis of the regular twin paradox:

- If you are seeing images sent by direct light path (a la a webcam), the turnaround twin seed the stay at home clock running fast for half the trip, while the stay at home sees the traveling twin clock fast for much less than half the trip.

- There are an infinite number of way to try to separate Doppler or pure visual image from some hypothetical underlying time rate. As a result, there is no uniquely defensible answer. However, any method of pairing the world lines for simultaneity (so the assignment is a continuous monotonic function of position on each world line, and pairs events with space like separation) has the following features:

-- there will be simultaneous times when one sees blue shift and the other redshift
-- the traveler will interpret the time rate of the stay at home as faster, on average, than theirs.

Note that the simultaneity match up that is based on instantly co-moving inertial observers is the so called Fermi-Normal coordinates. For this, if you smooth the turnaround even the tiniest bit, then during the turnaround, the turnaround twin interprets the stay at home time rate as extremely fast.

The circular twin paradox has all the same essential elements as above if signals pass through empty space. It is also essentially the same for any signal path that isn't like GSahi's (IMO absurd) case of communication only through a mirrored tube matching the path, and all signals only travel one way, thus forcing light to behave as if there is no turnaround. For example, if the webcams communicate via a space platform placed way over the pole (while one twin travels rapidly around the equator and the other stays put), then the traveling twill will see the the stay at home always fast, and the stay at home will see the traveler always slow. This is consequence of the fact that this particular signal path length is constant - IMO the most informative because you are then purely comparing time rate at fixed delay, without effects from varying length of signal path.
 
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  • #185
Austin0 said:
Does this mean you have changed your mind about your conclusions regarding the direct communication scenario. either a transparent Earth or no earth?
I think you have me confused with PAllen. I never made any argument about a transparent Earth or about no earth.

But since GAsahi has gotten himself banned and will not be able to answer my question, I will assume that my assessment of his position is accurate. I was then going to point out that even though in the typical twins paradox scenario (where the traveler goes away in a straight line and then turns around and comes back) and they both observe the same redshift at the start of the trip and the same blueshift at the end of the trip, the percentage of time that they observe redshift and blueshift is not the same. The traveler observes redshift for exactly the first half of the trip and blueshift for exactly the second half of the trip whereas the stay-at-home observer observes redshift for just about the entire time the traveler is gone and only observes blueshift during the very end of the trip. Therefore we can say that for both observers during most of the second half of their observations, the non-inertial traveler observes blueshift and the inertial observer observes redshift. I finally got GAsahi to admit that their observations of redshift are different in post #164 but then communication stopped:
GAsahi said:
Of course that the observer in circular motion is not seeing the same amount of redshift, since he isn't perfectly equivalent to the inertial observer. The point is that BOTH observers measure redshift. Please don't try to move goalposts, I am very good at detecting such attempts.

You should have stopped while you were still ahead, the point is (and has always been) that your statement that one observer measures redshift and the other one measures blueshift is false.
He was only able to defend his stance of both observers measuring the same redshift by focusing only on the first half of their observations and pretending that the second half of the trip didn't exist.

I have no idea whether GAsahi would admit to this explanation but if he would, I was then going to point out that each trip around the Earth is an individual twin paradox scenario and we could analyze it once and then multiply it by some huge number to get the full effect. Unfortunately, when GAsahi insists on using his fiberoptic link in his one-way scheme, he makes it impossible to link multiple trips together because it isn't over when the traveler gets back to his starting point.

The nice thing about the usual twin paradox scenario is that since the traveler spends most of his time inertial, the observed redshifts and blueshifts are steady and easy to analyze but when the traveler takes a circular path, they are continually changing for line of sight observations which of course is not possible in this scenario so the whole issue is how do we provide a communication scheme that will allow continual communication. I proposed in post #31 radio links via antenna(s) and stated that an antenna at or above the Earth's pole would allow for steady Dopplers, steady redshift Doppler observed by the son and steady blueshift Doppler observed by the father but even if they weren't steady because of the antenna locations, they would average out the same. Unfortunately, GAsahi made several wrong statements regarding this and now we will never have a chance to iron them out.

But rest assured, I have not changed my mind about anything on this thread.
 
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  • #186
PAllen said:
Actually, I disagree with this analysis of the regular twin paradox:
I was composing my previous response while you were posting yours but I think you can see that we are in complete agreement as long as you understand that I was framing the twin paradox as a paradox which states that both twins always determine the other twin as being time dilated. This of course works fine for the stay-at-home twin since he remains inertial but the traveling twin requires two separate inertial reference frames in order to create the apparent paradox. I just thought that this might be the point that GAsahi was making but now we'll never know.
 
  • #187
ghwellsjr said:
But rest assured, I have not changed my mind about anything on this thread.
glad to hear it


ghwellsjr said:
I think you have me confused with PAllen. I never made any argument about a transparent Earth or about no earth.


Originally Posted by ghwellsjr View Post


But in this thread, the stationary clock is not at the pole but at the equator which complicates things. For one, each clock will only be able to see the other one during a small portion of the time when they are close together. During this brief period of time, you can approximate the relative motion as mutual and they each see the other ones clock as ticking faster while approaching then slower while retreating but the time dilation, based on two different approximately inertial frames for each clock will determine that the other clock is ticking much more slowly (I presume this is what HallsofIvy meant in post #21). And if they could see through the earth, they would each continue to see the other ones clock fluctuating in its tick rate, but on average, the zipping clock would see the station clock as going faster than its own and the station clock would see the zipping clock as going slower than its own. That's why I said in post #20 "on average".

Actually it was this that got me interested in the scenario. And though both PAllen and myself simply eliminated the superfluous Earth the concept clearly came from you.


ghwellsjr said:
But since GAsahi has gotten himself banned and will not be able to answer my question,
That's too bad,how could that happen? :confused:


ghwellsjr said:
I will assume that my assessment of his position is accurate. I was then going to point out that even though in the typical twins paradox scenario (where the traveler goes away in a straight line and then turns around and comes back) and they both observe the same redshift at the start of the trip and at the end of the trip, the percentage of time that they observe redshift and blueshift is not the same. The traveler observes redshift for exactly the first half the trip and blueshift for exactly the second half of the trip whereas the stay-at-home observer observes redshift for just about the entire time the traveler is gone and only observes blueshift during the very end of the trip. Therefore we can say that for both observers during most of the second half of their observations, the non-inertial traveler observes blueshift and inertial observer observes redshift. I finally got GAsahi to admit that their observations of redshift are different in post #164 but then communication stopped:
Yes this is clear. An alternative but equivalent perspective is:

Exactly half the signals received by the inertial twin are redshifted.
Whereas the traveler only receives redshifted signals prior to turnaround which is necessarily much less than half the total number of signals for the full course.I.e. more blueshifted signals

Exactly how did you derive the gamma factor from these relationships?

How did you turn these percentages into elapsed times?
 
  • #188
ghwellsjr said:
I was composing my previous response while you were posting yours but I think you can see that we are in complete agreement as long as you understand that I was framing the twin paradox as a paradox which states that both twins always determine the other twin as being time dilated. This of course works fine for the stay-at-home twin since he remains inertial but the traveling twin requires two separate inertial reference frames in order to create the apparent paradox. I just thought that this might be the point that GAsahi was making but now we'll never know.

Fine, yes.
 
  • #189
Austin0 said:
ghwellsjr said:
I think you have me confused with PAllen. I never made any argument about a transparent Earth or about no earth.
ghwellsjr said:
But in this thread, the stationary clock is not at the pole but at the equator which complicates things. For one, each clock will only be able to see the other one during a small portion of the time when they are close together. During this brief period of time, you can approximate the relative motion as mutual and they each see the other ones clock as ticking faster while approaching then slower while retreating but the time dilation, based on two different approximately inertial frames for each clock will determine that the other clock is ticking much more slowly (I presume this is what HallsofIvy meant in post #21). And if they could see through the earth, they would each continue to see the other ones clock fluctuating in its tick rate, but on average, the zipping clock would see the station clock as going faster than its own and the station clock would see the zipping clock as going slower than its own. That's why I said in post #20 "on average".
Actually it was this that got me interested in the scenario. And though both PAllen and myself simply eliminated the superfluous Earth the concept clearly came from you.
I wasn't arguing for a transparent Earth here, in fact I continued with:
ghwellsjr said:
I believe uniqueland wanted to avoid all these complications, especially of not being able to see the other one during the entire orbit and so he introduced a couple webcams. Now it will depend on where the mutual antenna is located as to how much fluctuation would be seen by each observer. I mentally put this antenna above the Earth's pole to eliminate any fluctuation but to be more general, I allowed for the antenna or antennas to be located anywhere and so I included "on average".
I was trying to show that even though they couldn't see through the earth, any kind of continuous radio transmission would provide the same average Doppler as seeing through the earth. And by average, I mean over each trip of the father around the earth.
Austin0 said:
ghwellsjr said:
I will assume that my assessment of his position is accurate. I was then going to point out that even though in the typical twins paradox scenario (where the traveler goes away in a straight line and then turns around and comes back) and they both observe the same redshift at the start of the trip and the same blueshift at the end of the trip, the percentage of time that they observe redshift and blueshift is not the same. The traveler observes redshift for exactly the first half of the trip and blueshift for exactly the second half of the trip whereas the stay-at-home observer observes redshift for just about the entire time the traveler is gone and only observes blueshift during the very end of the trip. Therefore we can say that for both observers during most of the second half of their observations, the non-inertial traveler observes blueshift and the inertial observer observes redshift. I finally got GAsahi to admit that their observations of redshift are different in post #164 but then communication stopped:
Yes this is clear. An alternative but equivalent perspective is:

Exactly half the signals received by the inertial twin are redshifted.
Whereas the traveler only receives redshifted signals prior to turnaround which is necessarily much less than half the total number of signals for the full course.I.e. more blueshifted signals
This is not an alternative perspective nor is it equivalent. The non-inertial traveler sees redshift exactly the first half the trip and blueshift the last half. The inertial twin sees redshift almost all the time with just a little blueshift at the very end.

Note that we are not concerned with the number of signals but with the interval of time as measured by each observer's clock.
Austin0 said:
Exactly how did you derive the gamma factor from these relationships?
I thought I explained this in post #52 using your example. If you don't understand it, could you pinpoint the area of confusion?
Austin0 said:
How did you turn these percentages into elapsed times?
I'm not sure what you are looking for here. I really thought you had a good handle on this from your example and I'm afraid I might be stating the obvious. The easiest way would be to look at the traveling twin who sees redshift half the time and blueshift the other half. You could take the formula for Relativistic Doppler factor based on speed and average it with its reciprocal and that will give you gamma. This represents the ratio of the elapsed time of the inertial twin to the elapsed time of the traveling twin for any given speed. Is that the sort of thing you were looking for?
 
  • #190
If there is no absolute movement, moving at close to light speed has no meaning. If you are moving at the speed of light relatively to a reference frame, than anything moving with that frame is moving at the speed of light relatively to your frame. So, measured on the Earth frame you're not "ageing"; but measured on your frame your son is not "ageing" either.

Let's now imagine that your son embarks on a similar train which synchronizes his march with yours. You are both now on the same frame (yours) and your son should still be 1 year old (because in your frame no time has passed). But than he should go on a process of de-ageing, like Mr. Button, because when he embarked he was 30!??!?

Speculative discussions just lead to nowhere!

Let's look at an everyday experiment: satellite clocks.

Satellite clocks slow down relatively to Earth clocks (in fact they speed up because the effect of distance to the centre of mass (GR) is stronger at satellite altitudes). The slow down effect is quite correctly calculated by SR formulas.

This could mean that clocks slow down due to relative movement; but then, for this to be consistent, Earth clocks would slow down when measured by the satellite observer (the on-board computer).
But, everyday experience shows that this is not the case: for the on-board computer, Earth clocks also speed up due to relative movement, not slowed down.

This demonstrates that the slow down is not an effect of relative movement but only of movement relatively to the Dominant Mass. Dominant Mass (in this case Earth's mass) plays a dominant role: clocks slow down with their speed relatively to it and speed up with distance to its centre.

This is a point of view totally compatible with SR as it is applied: always relatively to Earth's reference frame. But, when applied to the satellite reference frame, it gives totally different results. Contrary to SR, this will predict that satellite clocks will slow down even on its own frame; like they really do.

Divirtam-se
Heitor

PS: The term "dominant mass" is not very precise and scientific; it's just a simplification not to deviate the discussion from it's propose: speed of light and time dilation.
 
  • #191
Simplyh said:
This demonstrates that the slow down is not an effect of relative movement but only of movement relatively to the Dominant Mass. Dominant Mass (in this case Earth's mass) plays a dominant role: clocks slow down with their speed relatively to it and speed up with distance to its centre.
This is incorrect. If we treat this purely as a question in Special Relativity and ignore gravitational effects, the explanation is that the satellite is following a circular path and therefore accelerating: it is not permanently at rest in an inertial frame so you cannot apply results that were derived for inertial frames.
 
  • #192
Dr. Greg
Lots of effects are taken to correct GPS systems everyday: one of them is SR relative movement; another, GR distance to the centre of mass or the weaker gravitational field if you want.

I think it was quite clear that I was referring to those two effects ignoring all others. If I was not clear I apologize.

Regarding the SR effect it is not at all relative: it's the same for the observer on the clock's frame (the on-board computer) and the observer on Earth frame. The acceleration effect is something else.

Diverte-te
Heitor
 
  • #193
Simplyh said:
If there is no absolute movement, moving at close to light speed has no meaning.
There is no absolute movement but we can meaningfully define moving at any speed up to the speed of light.
Simplyh said:
If you are moving at the speed of light relatively to a reference frame, than anything moving with that frame is moving at the speed of light relatively to your frame.
We cannot move at the speed of light and we cannot have one frame moving at the speed of light relatively to another frame.
Simplyh said:
So, measured on the Earth frame you're not "ageing"; but measured on your frame your son is not "ageing" either.
Of course the father is ageing, but at a very slow rate in the Earth frame due to his very high speed. And of course the son is aging at a very fast rate according to the father's (non-inertial) frame.
Simplyh said:
Let's now imagine that your son embarks on a similar train which synchronizes his march with yours. You are both now on the same frame (yours) and your son should still be 1 year old (because in your frame no time has passed).
Not "no time", just very little time. And the son was 5 at the start of this scenario, not 1. And just like the father, the son also would age a little bit if he took the trip but he didn't, so why introduce this totally irrelevant twist?
Simplyh said:
But than he should go on a process of de-ageing, like Mr. Button, because when he embarked he was 30!??!?
De-ageing?? What in the world does that mean? And the father was 35, not 30, when he got on the train. The trip lasted for 30 years. You're getting the details mixed up. The whole point of this thread is that at the start, the son is 5 and the father is 35. At the end, they are the same age, slightly over 35. But that requires that the son remain stationary and not take the trip. Since you disagree with this, you need to learn Special Relativity, not be trying to teach it.
Simplyh said:
Speculative discussions just lead to nowhere!
Then why are you doing it?
Simplyh said:
Let's look at an everyday experiment: satellite clocks.

Satellite clocks slow down relatively to Earth clocks (in fact they speed up because the effect of distance to the centre of mass (GR) is stronger at satellite altitudes). The slow down effect is quite correctly calculated by SR formulas.

This could mean that clocks slow down due to relative movement; but then, for this to be consistent, Earth clocks would slow down when measured by the satellite observer (the on-board computer).
But, everyday experience shows that this is not the case: for the on-board computer, Earth clocks also speed up due to relative movement, not slowed down.

This demonstrates that the slow down is not an effect of relative movement but only of movement relatively to the Dominant Mass. Dominant Mass (in this case Earth's mass) plays a dominant role: clocks slow down with their speed relatively to it and speed up with distance to its centre.

This is a point of view totally compatible with SR as it is applied: always relatively to Earth's reference frame. But, when applied to the satellite reference frame, it gives totally different results. Contrary to SR, this will predict that satellite clocks will slow down even on its own frame; like they really do.

Divirtam-se
Heitor

PS: The term "dominant mass" is not very precise and scientific; it's just a simplification not to deviate the discussion from it's propose: speed of light and time dilation.
In this thread, we have ignored the effects of gravity so why introduce it now? Your point of view sounds speculative to me if it is contrary to SR. The tick rate of a clock is dependent only on its speed within a frame. The Earth clocks are stationary and tick at the same rate as any clock used to define co-ordinate time within the frame. The moving clocks tick at a slower rate. The effect is not reciprocal because the moving clocks are not stationary in any inertial frame. All frames (inertial or non-inertial) will agree that the clocks moving in a circle will tick slower than the stationary inertial clocks. And they will agree that the moving clocks will observe that the stationary clocks are ticking faster than they are. That is one of the major points of this thread. I'm sorry that you disagree with this and want to come up with your own, incorrect point of view.
 
  • #194
Simplyh said:
If there is no absolute movement, moving at close to light speed has no meaning.

No *absolute* meaning, yes, because you have to refer the motion to something. But that's not the same as "no meaning", period.

Simplyh said:
If you are moving at the speed of light relatively to a reference frame, than anything moving with that frame is moving at the speed of light relatively to your frame. So, measured on the Earth frame you're not "ageing"; but measured on your frame your son is not "ageing" either.

As ghwellsjr pointed out, massive objects can't move at the speed of light; you should rephrase this to say "if you are moving at close to the speed of light..." etc. With that correction, yes, the "rate of aging" is symmetric, your son will seem to you to age slower, and you will seem to your son to age slower. But that's dependent on the fact that your relative motion doesn't change; if it changes, things get more complicated, as in the various "twin paradox" type scenarios, one of which you go on to propose:

Simplyh said:
Let's now imagine that your son embarks on a similar train which synchronizes his march with yours. You are both now on the same frame (yours) and your son should still be 1 year old (because in your frame no time has passed). But than he should go on a process of de-ageing, like Mr. Button, because when he embarked he was 30!??!?

This needs to be corrected as above; but even with the corrections, you are missing a critical element: the relativity of simultaneity. You say "when he embarked he was 30", but "when he embarked" is frame-dependent. You need to step back and carefully define the actual events involved and which frame you are defining them relative to (yours or your son's).
 
  • #195
PeterDonis said:
As ghwellsjr pointed out, massive objects can't move at the speed of light; you should rephrase this to say "if you are moving at close to the speed of light..." etc. With that correction, yes, the "rate of aging" is symmetric, your son will seem to you to age slower, and you will seem to your son to age slower. But that's dependent on the fact that your relative motion doesn't change; if it changes, things get more complicated, as in the various "twin paradox" type scenarios, one of which you go on to propose:
Simplyh said:
Let's now imagine that your son embarks on a similar train which synchronizes his march with yours. You are both now on the same frame (yours) and your son should still be 1 year old (because in your frame no time has passed). But than he should go on a process of de-ageing, like Mr. Button, because when he embarked he was 30!??!?
I'm afraid Simplyh's proposal was not a "twin paradox" type scenario. The original scenario proposed by the OP in post #1 was a "twin paradox" scenario repeated over and over again many times and with each loop of the father, the average rate of aging and thus the accumulated age difference is not symmetric. That's what this whole thread is about and I don't understand why it has generated so much confusion. It should have ended at post #23.
 
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  • #196
ghwellsjr said:
I'm afraid Simplyh's proposal was not a "twin paradox" type scenario.

I may well have been misunderstanding what he was proposing. I agree with you that the OP question is simple, and your analysis looks correct to me.
 
  • #197
"There is no absolute movement but we can meaningfully define moving at any speed up to the speed of light."

How can you do that unless you refer your movement to a reference frame arbitraryly chosen to be at rest? Let's call A the "moving" frame and B the frame at rest. Can't we switch the frames? Can't we call A the rest frame and B the moving frame. If A and B are moving relatively at |.5c|, then, for the observer in A, B is moving at |.5c| and he is at rest; but for the observer in B, he is at rest and A is moving at |.5c|.

So, speaking about speed has no meaning unless you refer it to a frame which you arbitraryly choosed to be at rest: we can not say "he's speeding at .5c"; we can only say: "he's speeding at .5c relatively to ... (whatever)". So the father is speeding at close to c relatively to the .son, but the son is also speeding at close to c, relatively to his father

Of course a circular movement is an accelerated movement, but that was not the point. If we were talking about acceleration, then we should not have said that the father was traveling at the speed of light (or close, not to cause nausea on the very delicate stomachs of some participants); we should have spoken about acceleration and its effect on the man's ageing, not about speed of light! If it was only about acceleration due to circular movement around the Earth, the calculations would not be, at all, the ones we were talking about and the guy would have arrived very close to 75; only the effect of relative speed could have make him stay 35. So, let's ignore the acceleration effect which is negligible compared to the speed effect. By the way, effects can be isolated and added: in this case the speed effect would have made the man stay practically at the same age and the acceleration would have made him ageing only an a bit more slowly.

Choosing the father's frame as to be at rest, the son is moving at close to the speed of light and, as so, ageing very slowly. For the father his son will always be 5 (isn't that true for all fathers?). So, if instead of puting the father meeting the son, we make the son embark on a train similar to his father's, then he should arrive to his father's frame at the age of 5. But as he was 35 (not 30, thanks for reminding me of that very, very important detail) when he embarked (on the Earth frame) how could he have gone backwards on his age (not de-ageing, forgive my bad, bad English!).

One can make speculative thinking just to prove it is absurd.

Divirtam-se

PS: tomorrow I'll talk about satellites.
 
  • #198
Simplyh said:
"There is no absolute movement but we can meaningfully define moving at any speed up to the speed of light."

How can you do that unless you refer your movement to a reference frame arbitraryly chosen to be at rest?
Reference frames are arbitrarily chosen, but not necessarily at rest, the state of motion of the frame itself is immaterial as you point out in your next sentence:
Simplyh said:
Let's call A the "moving" frame and B the frame at rest. Can't we switch the frames? Can't we call A the rest frame and B the moving frame. If A and B are moving relatively at |.5c|, then, for the observer in A, B is moving at |.5c| and he is at rest; but for the observer in B, he is at rest and A is moving at |.5c|.
Both observers are in both frames. Sometimes we refer to a frame as being an observer's frame, meaning that he is at rest in that frame, but he does not hold exclusive rights to the frame, all other observers and objects are in the frame with him, it's just that they are moving in his rest frame.
Simplyh said:
So, speaking about speed has no meaning unless you refer it to a frame which you arbitraryly choosed to be at rest: we can not say "he's speeding at .5c"; we can only say: "he's speeding at .5c relatively to ... (whatever)".
That "whatever" is always a stated frame in Special Relativity, arbitrarily chosen, as you say.
Simplyh said:
So the father is speeding at close to c relatively to the .son, but the son is also speeding at close to c, relatively to his father
But there's a difference because the son is stationary in the Earth's inertial frame and the father is not stationary in any inertial frame. It gets really messy if you want to consider the son's motion in the father's non-inertial frame.
Simplyh said:
Of course a circular movement is an accelerated movement, but that was not the point. If we were talking about acceleration, then we should not have said that the father was traveling at the speed of light (or close, not to cause nausea on the very delicate stomachs of some participants); we should have spoken about acceleration and its effect on the man's ageing, not about speed of light! If it was only about acceleration due to circular movement around the Earth, the calculations would not be, at all, the ones we were talking about and the guy would have arrived very close to 75; only the effect of relative speed could have make him stay 35. So, let's ignore the acceleration effect which is negligible compared to the speed effect. By the way, effects can be isolated and added: in this case the speed effect would have made the man stay practically at the same age and the acceleration would have made him ageing only an a bit more slowly.
There is no ageing effect caused by acceleration (unless it results in a change in speed). In this scenario, the speed of the father is constant. All his acceleration causes him to move in a circle and does not change his speed. Only speed (relative to our chosen frame) effects time dilation.
Simplyh said:
Choosing the father's frame as to be at rest, the son is moving at close to the speed of light and, as so, ageing very slowly.
The father's rest frame is non-inertial which makes it very complex. If you want to carry this out, be my guest, but it cannot yield a different result than the son's rest frame.
Simplyh said:
For the father his son will always be 5 (isn't that true for all fathers?).
No, this cannot be. On every trip around the earth, the father is co-located with the son and every frame will agree about their accumulated age difference.
Simplyh said:
So, if instead of puting the father meeting the son, we make the son embark on a train similar to his father's, then he should arrive to his father's frame at the age of 5. But as he was 35 (not 30, thanks for reminding me of that very, very important detail) when he embarked (on the Earth frame) how could he have gone backwards on his age (not de-ageing, forgive my bad, bad English!).
Like I said before, if they are both traveling at the same speed (according to our arbitrarily chosen frame), then they will age the same but that's a new scenario having nothing to do with this thread. Why are you bringing it up?
Simplyh said:
One can make speculative thinking just to prove it is absurd.
Who's doing the speculative thinking?
Simplyh said:
Divirtam-se
English is required on this forum.
Simplyh said:
PS: tomorrow I'll talk about satellites.
You shouldn't bring up satellites on this thread. It's off topic, just like your scenario of putting the son on a train. You should start your own thread if you want to do that.
 
  • #199
ghwellsjr said:
There is no ageing effect caused by acceleration (unless it results in a change in speed). In this scenario, the speed of the father is constant. All his acceleration causes him to move in a circle and does not change his speed. Only speed (relative to our chosen frame) effects time dilation.
The father's circular (centripetal) acceleration doesn't affect the son's view that his father is ageing slowly (i.e., that his father's time is "dilated"), but it does affect the father's view about how fast his son is ageing. The father's centripetal acceleration causes him to say that his son is ageing very quickly (i.e., that his son's time is "contracted", or whatever term you want to use for the opposite of "dilated").
 
  • #200
GrammawSally said:
The father's circular (centripetal) acceleration doesn't affect the son's view that his father is ageing slowly (i.e., that his father's time is "dilated"), but it does affect the father's view about how fast his son is ageing. The father's centripetal acceleration causes him to say that his son is ageing very quickly (i.e., that his son's time is "contracted", or whatever term you want to use for the opposite of "dilated").
When we're talking about time dilation, we're talking about what we, not the observers in our scenario, see according to the frame that we have selected. There is no opposite of time dilation. What you are calling "contracted" is merely the normal co-ordinate time of the selected frame. It's only because the father's time is dilated in that frame that he perceives his son's ageing to be faster than his own.
 
  • #201
ghwellsjr said:
When we're talking about time dilation, we're talking about what we, not the observers in our scenario, see according to the frame that we have selected. There is no opposite of time dilation. What you are calling "contracted" is merely the normal co-ordinate time of the selected frame. It's only because the father's time is dilated in that frame that he perceives his son's ageing to be faster than his own.

Looked at from a metric point of view, the statement that there is only time dilation in an inertial frame is due to the expression of the flat space metric in such a frame. If one defines a metric for a non-inertial frame in any consistent way, you then have the possibility of both time dilation and time speed up. Specifically, the proper time between two events can be longer than the coordinate time difference between them in an accelerated frame. Of course, I agree there is little point in analyzing SR from that point of view except as a possible bridge to GR.
 
  • #202
PAllen said:
Looked at from a metric point of view, the statement that there is only time dilation in an inertial frame is due to the expression of the flat space metric in such a frame. If one defines a metric for a non-inertial frame in any consistent way, you then have the possibility of both time dilation and time speed up. Specifically, the proper time between two events can be longer than the coordinate time difference between them in an accelerated frame. Of course, I agree there is little point in analyzing SR from that point of view except as a possible bridge to GR.
I should have specified inertial frames. I know nothing about non-inertial frames and although I see a lot of talk about them, I don't recall anyone actually defining one and pointing out how it works. But as I pointed out to GAsahi in post #31:
ghwellsjr said:
GAsahi said:
Same difference, you are mixing up mutual time dilation (the way the two observers see each other moving) with the calculation of total elapsed time (the circular variant of the twins paradox).
You are mixing up time dilation (which is frame dependent and arbitrary) with what observers see (which is relativistic Doppler and not dependent on any arbitrarily selected frame).
time dilation (in any frame, inertial or non-inertial) has no bearing on what observers actually see and that is my point here.
 
  • #203
ghwellsjr said:
I should have specified inertial frames. I know nothing about non-inertial frames and although I see a lot of talk about them, I don't recall anyone actually defining one and pointing out how it works.

They do that on this web sight:

https://sites.google.com/site/cadoequation/cado-reference-frame
 
  • #204
ghwellsjr said:
I should have specified inertial frames. I know nothing about non-inertial frames and although I see a lot of talk about them, I don't recall anyone actually defining one and pointing out how it works. But as I pointed out to GAsahi in post #31:

time dilation (in any frame, inertial or non-inertial) has no bearing on what observers actually see and that is my point here.

Rindler coordinates represent a plausible coordinates for a uniformly accelerating observer in SR. Many books on GR discuss accelerated SR observers as a bridge to GR (IMO almost the only purpose that makes sense). MTW has a whole chapter on this. You can also check:

http://en.wikipedia.org/wiki/Rindler_coordinates

I completely agree with your last statement. What you see is just Doppler. Relative time rates for separated observers in some state of motion is an interpretation.
 
  • #205
PAllen said:
Specifically, the proper time between two events can be longer than the coordinate time difference between them in an accelerated frame. Of course, I agree there is little point in analyzing SR from that point of view except as a possible bridge to GR.

A person who is accelerating is still entitled to his own "point of view" about how other people are aging. Gravity or no gravity doesn't change that.
 
  • #206
GrammawSally said:
A person who is accelerating is still entitled to his own "point of view" about how other people are aging. Gravity or no gravity doesn't change that.
Accelerating or inertial, there is no "point of view" about how other remote people are aging. That's the whole point of relativity. Each different frame offers a different "point of view". If a person is inertial, and has been for a long time, and has some way of knowing what's actually going on with the other remote people because of prior agreement, and he chooses to use Special Relativity (Einstein's timing convention) and he chooses to select an inertial frame in which he is at rest, then we can say that there is a single "point of view" for that person about how the other people are aging. But if you want to talk about a person who is accelerating, even if all the other things I mentioned are still intact, there is no single standard way to define a frame that answers the issue of how the others are aging. Or to put it another way, two people who are accelerating together can come up with their own different "points of view" about how the others are aging. Do you agree with that?

But this thread is not about "points of view" in the context of what different people arbitrarily select as a Frame of Reference, it's about "points of view" in the context of what people actually see when looking at the image of another person. In that context, it is not arbitrary and there is only one answer or description and all Frames of Reference will yield the same answer. Since the son-earth inertial frame provides a simple way to calculate what both the father and son see of each other, the problem is solved without resorting to complicated non-inertial frames. But if that process appeals to you, I invite you to teach me how it is done.
 
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  • #207
Underwood said:
Have you actually done it for an example such as the one in this thread where the father is traveling around the Earth at a high speed and wants to determine the age of his son back in the train station at each point around the earth? I invite you to work it out and teach me how to do it. One trip around the Earth would be sufficient. And then after you do that, you need to show how that analysis determines what the father actually sees of the son's aging and vice versa, because that is what this thread is about.

Also, you should be aware that the CADO process is just one of an infinite number of ways to answer the unanswerable question of what is the Current Age of a Distant Object.
 
  • #208
ghwellsjr said:
Have you actually done it for an example such as the one in this thread where the father is traveling around the Earth at a high speed and wants to determine the age of his son back in the train station at each point around the earth? I invite you to work it out and teach me how to do it.

For two or three dimensional motion, the v * L term in the CADO equation becomes a dot product between the vectors v and L. Whenever v and L are perpendicular (which is the case in the circular scenario), that dot product is zero, and the CADO equation then says that the two twins will agree (during the entire circular motion) about the correspondence between their ages. I.e., they will both say that it is the home twin who is aging faster during the circular motion.
 
  • #209
GrammawSally said:
ghwellsjr said:
Have you actually done it for an example such as the one in this thread where the father is traveling around the Earth at a high speed and wants to determine the age of his son back in the train station at each point around the earth? I invite you to work it out and teach me how to do it.
For two or three dimensional motion, the v * L term in the CADO equation becomes a dot product between the vectors v and L. Whenever v and L are perpendicular (which is the case in the circular scenario), that dot product is zero, and the CADO equation then says that the two twins will agree (during the entire circular motion) about the correspondence between their ages. I.e., they will both say that it is the home twin who is aging faster during the circular motion.
Did you place the son at the center of the Earth instead of in the train station?
 
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  • #210
ghwellsjr said:
Did you place the son at the center of the Earth instead of in the train station?

Yes, I did. I hadn't looked at the original problem description for a while.

If the inertial home twin is located somewhere on the circular track, then the traveler can compute the dot product of v and L at any point on his circuit, and then use the CADO equation to compute the home twin's age at that instant in his (the traveler's) life.
 

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